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Page 1
Self ionization of water
Concentration of H
2
O ([H
2
O]) =
1000 / 18
55.5 M
1
?
Water is very weak electrolyte
H
2
O H
+
+ OH
–
55.5
55.5 – 10
–7
10
–7
10
–7
(at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
Constant
? K
w
= k
eq
[H
2
O] = [H
+
] [OH
–
]
? K
w
= [H
+
] [OH
–
],at 25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
(25
o
C) = 10
–14
?
Self ionization constant of water
let,
1
w
K
at temperature T
1
,
2
w
K
at temperature T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
–
T
1
R 2.303
H
K
K
log
1
2
, ? ? ? K T
w
? For calculation of ? :
H
2
O H
+
+ OH
–
c 0 0
c (1 – ?) c ? ? ??? ? ? ?? ?c ?
Page 2
Self ionization of water
Concentration of H
2
O ([H
2
O]) =
1000 / 18
55.5 M
1
?
Water is very weak electrolyte
H
2
O H
+
+ OH
–
55.5
55.5 – 10
–7
10
–7
10
–7
(at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
Constant
? K
w
= k
eq
[H
2
O] = [H
+
] [OH
–
]
? K
w
= [H
+
] [OH
–
],at 25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
(25
o
C) = 10
–14
?
Self ionization constant of water
let,
1
w
K
at temperature T
1
,
2
w
K
at temperature T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
–
T
1
R 2.303
H
K
K
log
1
2
, ? ? ? K T
w
? For calculation of ? :
H
2
O H
+
+ OH
–
c 0 0
c (1 – ?) c ? ? ??? ? ? ?? ?c ?
K
w
= c ? × c ?
?
c
k
w
? ? c = 5 5.56 M ?
w
–2
K 10 1.8 ? ? ?
?
pH
potent
?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH = – log [H
+
]
p OH = – log [OH
–
]
p K
w
= – log K
w
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
] [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
= 10
–14
K
w
= 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
5 L
pH = 3 ?
1 L
pH = ?
3 L
pH = 3 +
2 L
pH = 3 ?
5 L
pH = ?
? Concept 3: ? Concept 4:
2 L
pH = 3 +
8 L
HO
2
=
10 L
pH = ?
2 L
pH = 13 +
8 L
water =
10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate pH
? Concept 5:
pH = 3
+
pH = 4
=
pH = ?
Page 3
Self ionization of water
Concentration of H
2
O ([H
2
O]) =
1000 / 18
55.5 M
1
?
Water is very weak electrolyte
H
2
O H
+
+ OH
–
55.5
55.5 – 10
–7
10
–7
10
–7
(at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
Constant
? K
w
= k
eq
[H
2
O] = [H
+
] [OH
–
]
? K
w
= [H
+
] [OH
–
],at 25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
(25
o
C) = 10
–14
?
Self ionization constant of water
let,
1
w
K
at temperature T
1
,
2
w
K
at temperature T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
–
T
1
R 2.303
H
K
K
log
1
2
, ? ? ? K T
w
? For calculation of ? :
H
2
O H
+
+ OH
–
c 0 0
c (1 – ?) c ? ? ??? ? ? ?? ?c ?
K
w
= c ? × c ?
?
c
k
w
? ? c = 5 5.56 M ?
w
–2
K 10 1.8 ? ? ?
?
pH
potent
?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH = – log [H
+
]
p OH = – log [OH
–
]
p K
w
= – log K
w
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
] [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
= 10
–14
K
w
= 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
5 L
pH = 3 ?
1 L
pH = ?
3 L
pH = 3 +
2 L
pH = 3 ?
5 L
pH = ?
? Concept 3: ? Concept 4:
2 L
pH = 3 +
8 L
HO
2
=
10 L
pH = ?
2 L
pH = 13 +
8 L
water =
10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate pH
? Concept 5:
pH = 3
+
pH = 4
=
pH = ?
2 : 1
? Note : The final pH of solution after mixing two solution is in between the previous solution pH.
? Concept 6: ? Concept 7:
pH = 3
V
+ pH = 11
V
? pH = ?
2 V
pH = ?
10 M HCl
–8
? Note :
When [H
+
] > 10
–5
? [H
+
]
water
neglected
When [H
+
] < 10
–5
? [H
+
]
water
considered
H
2
O H
+
+ OH
–
55.56 10
–8
55.56 – x 10
–8
+ x x
? x (10
–8
+ x) = 10
–14
? x = .94 × 10
–7
? [H
+
] = 10
–8
+ .94 × 10
–7
= 1.04 × 10
–7
? pH = 7 – log (1.04)
? Concept 8:
pH = ?
10 M H Cl
2
? Dissociation of weak acid :
H
+
A
–
H
+
+ A
–
c
c (1– ?) c ? c ?
?
?
?
– 1
c
K
2
a
for weak acid ? ?< < 1 K
a
= c ?
2
[H
+
] = c ? ? =
a
K c , pH = – log
a
K c , for weak bases [H
+
] =
b
w
K c
K
?
=
] OH [
K
w
?
? Note :
(a) HA
1
c
K
a
1
||
HA
2
c
K
a
2
?
2
1
a
a
2
1
K
K
] H [
] H [
?
?
?
Page 4
Self ionization of water
Concentration of H
2
O ([H
2
O]) =
1000 / 18
55.5 M
1
?
Water is very weak electrolyte
H
2
O H
+
+ OH
–
55.5
55.5 – 10
–7
10
–7
10
–7
(at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
Constant
? K
w
= k
eq
[H
2
O] = [H
+
] [OH
–
]
? K
w
= [H
+
] [OH
–
],at 25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
(25
o
C) = 10
–14
?
Self ionization constant of water
let,
1
w
K
at temperature T
1
,
2
w
K
at temperature T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
–
T
1
R 2.303
H
K
K
log
1
2
, ? ? ? K T
w
? For calculation of ? :
H
2
O H
+
+ OH
–
c 0 0
c (1 – ?) c ? ? ??? ? ? ?? ?c ?
K
w
= c ? × c ?
?
c
k
w
? ? c = 5 5.56 M ?
w
–2
K 10 1.8 ? ? ?
?
pH
potent
?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH = – log [H
+
]
p OH = – log [OH
–
]
p K
w
= – log K
w
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
] [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
= 10
–14
K
w
= 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
5 L
pH = 3 ?
1 L
pH = ?
3 L
pH = 3 +
2 L
pH = 3 ?
5 L
pH = ?
? Concept 3: ? Concept 4:
2 L
pH = 3 +
8 L
HO
2
=
10 L
pH = ?
2 L
pH = 13 +
8 L
water =
10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate pH
? Concept 5:
pH = 3
+
pH = 4
=
pH = ?
2 : 1
? Note : The final pH of solution after mixing two solution is in between the previous solution pH.
? Concept 6: ? Concept 7:
pH = 3
V
+ pH = 11
V
? pH = ?
2 V
pH = ?
10 M HCl
–8
? Note :
When [H
+
] > 10
–5
? [H
+
]
water
neglected
When [H
+
] < 10
–5
? [H
+
]
water
considered
H
2
O H
+
+ OH
–
55.56 10
–8
55.56 – x 10
–8
+ x x
? x (10
–8
+ x) = 10
–14
? x = .94 × 10
–7
? [H
+
] = 10
–8
+ .94 × 10
–7
= 1.04 × 10
–7
? pH = 7 – log (1.04)
? Concept 8:
pH = ?
10 M H Cl
2
? Dissociation of weak acid :
H
+
A
–
H
+
+ A
–
c
c (1– ?) c ? c ?
?
?
?
– 1
c
K
2
a
for weak acid ? ?< < 1 K
a
= c ?
2
[H
+
] = c ? ? =
a
K c , pH = – log
a
K c , for weak bases [H
+
] =
b
w
K c
K
?
=
] OH [
K
w
?
? Note :
(a) HA
1
c
K
a
1
||
HA
2
c
K
a
2
?
2
1
a
a
2
1
K
K
] H [
] H [
?
?
?
? So HA
2
is stronger acid than HA
1
(b) HA
1
c
1
K
a
1
HA
2
c
2
K
a
2
2 2
1 1
2
1
Ka c
ka c
] H [
] H [
?
?
?
? The acidic strength of two weak acid can be compared by K
a
value only in the case when their concen-
tration are same. Otherwise the procedure of case (b) should be followed .
?
pH of mixture of two acids
?
? Case (I) (Strong acid + weak acid)
Question : HCl (0.1 M) + CH
3
COOH (0.2 M)
?
K
a
= 10
–5
CH
3
COOH CH
3
COO
–
+ H
+
t = 0 0.2 0 0. 1
t = t
eq
0.2 – x x 0.1 + x (x is very small)
?
?
?
0.2 x 0.1
? pH = – log (0.1) = 1 ? [CH
3
COO
–
] = ? ?
5 – –
3
5 –
10 2 ] COO [CH
2 . 0
1 . 0 x
10 ? ? ?
?
?
? Case (II-A) (weak acid + weak acid ) (k
as
are incomparable)
Question : HA
1
(0.1 M,
K
a
1
= 10
–5
)
HA
2
(0.2 M,
K
a
2
= 10
–9
)
HA
1
H
+
+ A
1
–
HA
2
H
+
+ A
2
–
0.1 0 0 0.2 0 0
0.1 – x x + y x 0.2 – y x + y y (
K
a
1
>>
K
a
2
?
? ? ?
?
?
? x > > y)
0.1 x x 0.2 x y
x and y are also very less
Page 5
Self ionization of water
Concentration of H
2
O ([H
2
O]) =
1000 / 18
55.5 M
1
?
Water is very weak electrolyte
H
2
O H
+
+ OH
–
55.5
55.5 – 10
–7
10
–7
10
–7
(at 25
o
C)
?
O] [H
] [OH ] [H
k
2
–
eq
?
?
Constant
? K
w
= k
eq
[H
2
O] = [H
+
] [OH
–
]
? K
w
= [H
+
] [OH
–
],at 25
o
C, [H
+
] = [OH
–
] = 10
–7
K
w
(25
o
C) = 10
–14
?
Self ionization constant of water
let,
1
w
K
at temperature T
1
,
2
w
K
at temperature T
2
?
?
?
?
?
?
? ?
?
2 1 w
w
T
1
–
T
1
R 2.303
H
K
K
log
1
2
, ? ? ? K T
w
? For calculation of ? :
H
2
O H
+
+ OH
–
c 0 0
c (1 – ?) c ? ? ??? ? ? ?? ?c ?
K
w
= c ? × c ?
?
c
k
w
? ? c = 5 5.56 M ?
w
–2
K 10 1.8 ? ? ?
?
pH
potent
?
strength
H ydrogen ion
(pH s cale is given by S orension)
? pH = – log [H
+
]
p OH = – log [OH
–
]
p K
w
= – log K
w
at 25°C
Acidic basic
0
7 14
Neutral
p scale
H
at any temperature :
[H
+
] = [OH
–
]
[H
+
] [OH
–
] = K
w
? – log [H
+
] – log [OH
–
] = – log [K
w
]
? pH + pOH = p K
w
at 25
o
C at 80
o
C
K
w
= 10
–14
K
w
= 10
–12
pH + pOH = 14 pH + pOH = 12
Some concept regarding pH calculation
? Concept 1 : ? Concept 2:
5 L
pH = 3 ?
1 L
pH = ?
3 L
pH = 3 +
2 L
pH = 3 ?
5 L
pH = ?
? Concept 3: ? Concept 4:
2 L
pH = 3 +
8 L
HO
2
=
10 L
pH = ?
2 L
pH = 13 +
8 L
water =
10 L
pH = ?
? Note :
7
pH
(i) Calculate pOH
(ii) pH = 14- pOH
Calculate pH
? Concept 5:
pH = 3
+
pH = 4
=
pH = ?
2 : 1
? Note : The final pH of solution after mixing two solution is in between the previous solution pH.
? Concept 6: ? Concept 7:
pH = 3
V
+ pH = 11
V
? pH = ?
2 V
pH = ?
10 M HCl
–8
? Note :
When [H
+
] > 10
–5
? [H
+
]
water
neglected
When [H
+
] < 10
–5
? [H
+
]
water
considered
H
2
O H
+
+ OH
–
55.56 10
–8
55.56 – x 10
–8
+ x x
? x (10
–8
+ x) = 10
–14
? x = .94 × 10
–7
? [H
+
] = 10
–8
+ .94 × 10
–7
= 1.04 × 10
–7
? pH = 7 – log (1.04)
? Concept 8:
pH = ?
10 M H Cl
2
? Dissociation of weak acid :
H
+
A
–
H
+
+ A
–
c
c (1– ?) c ? c ?
?
?
?
– 1
c
K
2
a
for weak acid ? ?< < 1 K
a
= c ?
2
[H
+
] = c ? ? =
a
K c , pH = – log
a
K c , for weak bases [H
+
] =
b
w
K c
K
?
=
] OH [
K
w
?
? Note :
(a) HA
1
c
K
a
1
||
HA
2
c
K
a
2
?
2
1
a
a
2
1
K
K
] H [
] H [
?
?
?
? So HA
2
is stronger acid than HA
1
(b) HA
1
c
1
K
a
1
HA
2
c
2
K
a
2
2 2
1 1
2
1
Ka c
ka c
] H [
] H [
?
?
?
? The acidic strength of two weak acid can be compared by K
a
value only in the case when their concen-
tration are same. Otherwise the procedure of case (b) should be followed .
?
pH of mixture of two acids
?
? Case (I) (Strong acid + weak acid)
Question : HCl (0.1 M) + CH
3
COOH (0.2 M)
?
K
a
= 10
–5
CH
3
COOH CH
3
COO
–
+ H
+
t = 0 0.2 0 0. 1
t = t
eq
0.2 – x x 0.1 + x (x is very small)
?
?
?
0.2 x 0.1
? pH = – log (0.1) = 1 ? [CH
3
COO
–
] = ? ?
5 – –
3
5 –
10 2 ] COO [CH
2 . 0
1 . 0 x
10 ? ? ?
?
?
? Case (II-A) (weak acid + weak acid ) (k
as
are incomparable)
Question : HA
1
(0.1 M,
K
a
1
= 10
–5
)
HA
2
(0.2 M,
K
a
2
= 10
–9
)
HA
1
H
+
+ A
1
–
HA
2
H
+
+ A
2
–
0.1 0 0 0.2 0 0
0.1 – x x + y x 0.2 – y x + y y (
K
a
1
>>
K
a
2
?
? ? ?
?
?
? x > > y)
0.1 x x 0.2 x y
x and y are also very less
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53
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1 –
5 –
10
x . x
10 ?
[H
+
] = x = 10
–3
, ? pH = 3 [A
1
–
] = [A
2
–
] =
? Case (II-B) (weak acid + weak acid ) (k
as
are comparable)
?Question : HA
1
(0.1)
K
a
1
= 10
–5
HA
2
(0.2)
K
a
2
= 10
–6
HA
1
H
+
+ A
1
–
HA
2
H
+
+ A
2
–
0.1 0 0 0.2 0 0
0.1 – x x + y x 0.2 – y x + y y
0.1 x + y x 0.2 x + y y
? now
0.1
y) (x x
K
1
a
?
?
?
1
a
c
y) (x x
K
1
?
?
,
2
a
c
y y) (x
K
2
?
?
?
?
2 1
a 2 a 1
K c K c y x ] [H ? ? ? ?
?
? weak polyprotic acids or bases
(H
2
S, H
3
PO
4
, NH
2
– NH
2
, H
2
CO
3
)
(a) H
2
S H
+
+ HS
–
K
a
1
HS
–
H
+
+ S
2–
K
a
2
(b) NH
2
– NH
2
+ H
+
NH
2
– NH
3
+
1
b
K
NH
2
– NH
3
+
+ H
+
H N – H N
3 3
? ?
2
b
K
Cases
? Case I : H
2
S (0.1 M
K
a
1
= 10
–4
,
K
a
2
= 10
–9
)
? H
2
S
H
+
+ H S
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FAQs on Self Ionization of Water and Buffer - Chemistry for JEE Main & Advanced
| 1. What is the self-ionization of water? |  |
Ans. The self-ionization of water refers to the process in which water molecules can spontaneously dissociate into ions. This occurs when a water molecule donates a proton (H+) to another water molecule, forming a hydronium ion (H3O+) and a hydroxide ion (OH-). The self-ionization of water can be represented by the equation: 2H2O ⇌ H3O+ + OH-.
| 2. Why is the self-ionization of water important in the context of buffers? | |
Ans. The self-ionization of water is important in the context of buffers because it plays a crucial role in maintaining the pH stability of a buffer solution. A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) and resists changes in pH when small amounts of acid or base are added. The self-ionization of water provides a source of hydrogen ions (H+) and hydroxide ions (OH-) that can interact with the weak acid or base in the buffer solution, helping to maintain a stable pH.
| 3. How does the self-ionization of water affect the pH of a buffer solution? | |
Ans. The self-ionization of water affects the pH of a buffer solution by providing a source of hydrogen ions (H+) and hydroxide ions (OH-) that can influence the concentration of these ions in the solution. In a buffer solution, the weak acid or base reacts with these ions, resulting in a dynamic equilibrium that resists changes in pH. If the concentration of hydrogen ions increases (due to the self-ionization of water), the weak base component of the buffer will react with them, shifting the equilibrium towards the weak acid form and maintaining the pH. Similarly, if the concentration of hydroxide ions increases, the weak acid component of the buffer will react with them, shifting the equilibrium towards the weak base form and maintaining the pH.
| 4. Can the self-ionization of water be affected by external factors? | |
Ans. Yes, the self-ionization of water can be affected by external factors such as temperature and pressure. An increase in temperature generally leads to an increase in the self-ionization of water, resulting in a higher concentration of hydrogen ions (H+) and hydroxide ions (OH-). Conversely, a decrease in temperature decreases the self-ionization. Changes in pressure, on the other hand, have a negligible effect on the self-ionization of water.
| 5. How is the pH of a buffer solution affected by the concentration of the weak acid and its conjugate base? | |
Ans. The pH of a buffer solution is affected by the concentration of the weak acid and its conjugate base through the Henderson-Hasselbalch equation. This equation relates the pH, the pKa (acid dissociation constant) of the weak acid, and the ratio of the concentrations of the weak acid and its conjugate base in the buffer solution. As per the Henderson-Hasselbalch equation, at a given temperature, pH = pKa + log([conjugate base]/[weak acid]). Therefore, by altering the concentrations of the weak acid and its conjugate base, the pH of the buffer solution can be adjusted. Increasing the concentration of the weak acid relative to the conjugate base will lower the pH, while increasing the concentration of the conjugate base relative to the weak acid will increase the pH.
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